a(n) is also the number of conjugacy classes in the symmetric group S_n (and the number of irreducible representations of S_n).

Also the number of rooted trees with n+1 nodes and height at most 2.

Coincides with the sequence of numbers of nilpotent conjugacy classes in the Lie algebras gl(n). A006950, A015128 and this sequence together cover the nilpotent conjugacy classes in the classical A,B,C,D series of Lie algebras. - Alexander Elashvili, Sep 08 2003

Let P(z) := Sum_{j>=0} b_j z^j, b_0 != 0. Then 1/P(z) = Sum_{j>=0} c_j z^j, where the c_j must be computed from the infinite triangular system b_0 c_0 = 1, b_0 c_1 + b_1 c_0 = 0 and so on (Cauchy products of the coefficients set to zero). The n-th partition number arises as the number of terms in the numerator of the expression for c_n: The coefficient c_n of the inverted power series is a fraction with b_0^(n+1) in the denominator and in its numerator having a(n) products of n coefficients b_i each. The partitions may be read off from the indices of the b_i. - Peter C. Heinig (algorithms(AT)gmx.de), Apr 09 2007

a(n) = the number of different ways to run up a staircase with n steps, taking steps of sizes 1, 2, 3, ... and r (r <= n), where the order is not important and there is no restriction on the number or the size of each step taken. - Mohammad K. Azarian, May 21 2008

a(n) is the number of distinct degree sequences over all free trees having n + 2 nodes. Take a partition of the integer n, add 1 to each part and append as many 1's as needed so that the total is 2n + 2. Now we have a degree sequence of a tree with n + 2 nodes. Example: The partition 3 + 2 + 1 = 6 corresponds to the degree sequence {4, 3, 2, 1, 1, 1, 1, 1} of a tree with 8 vertices. - Geoffrey Critzer, Apr 16 2011

Partitions of n that contain a part p are the partitions of n - p. Thus, number of partitions of m*n - r that include k*n as a part is A000041(h*n-r), where h = m - k >= 0, n >= 2, 0 <= r < n; see A111295 as an example. - Clark Kimberling, Mar 03 2014

a(n) is the number of compositions of n into positive parts avoiding the pattern [1, 2]. - Bob Selcoe, Jul 08 2014

Conjecture: For any j there exists k such that all primes p <= A000040(j) are factors of one or more a(n) <= a(k). Growth of this coverage is slow and irregular. k = 1067 covers the first 102 primes, thus slower than A000027. - Richard R. Forberg, Dec 08 2014

a(n) is the number of nilpotent conjugacy classes in the order-preserving, order-decreasing and (order-preserving and order-decreasing) injective transformation semigroups. - Ugbene Ifeanyichukwu, Jun 03 2015

Define a segmented partition a(n,k, <s(1)..s(j)>) to be a partition of n with exactly k parts, with s(j) parts t(j) identical to each other and distinct from all the other parts. Note that n>=k, j<=k, 0<=s(j)<=k, s(1)t(1) + ... + s(j)t(j) = n and s(1) + ... + s(j) = k. Then there are up to a(k) segmented partitions of n with exactly k parts.

a(n) - a(n-1) - a(n-2) + a(n-5) + a(n-7) - a(n-12) - a(n-15) + ... = 0, where the sum is over n-k and k is a generalized pentagonal number (A001318) <= n and the sign of the k-th term is (-1)^([(k+1)/2]). See A001318 for a good way to remember this!

a(n) = (1/n) * Sum_{k=0..n-1} sigma(n-k)*a(k), where sigma(k) is the sum of divisors of k (A000203).

where d_q = (-1)^(m+1) if q = m(3m-1)/2 = p_m, the m-th generalized pentagonal number (A001318), otherwise d_q = 0. Note that the 1's run along the diagonal and the -1's are on the superdiagonal. The (n-1) row (not written) would end with ... 1 -1. (End)

F*(4/5) = 1/2+3/2/sqrt(5)-sqrt(1/2*(1+3/sqrt(5))) to a precision of 28 digits. These are the only values found for a/b when a/b is from F60, Farey fractions up to 60. The number for F*(4/5) is one of the real roots of 25*x^4 - 50*x^3 - 10*x^2 - 10*x + 1. Note here the exponent (n+1) compared to the standard notation with n starting at 0. - Simon Plouffe, Feb 23 2011

a(n-3) = (1/2)*Sum_{parts k in all partitions in P(3,n)} phi(k), where phi(k) is the Euler totient function (see A000010). Using this result and Merten's theorem on the average order of the phi function, we can find an approximate 3-term recurrence for the partition function:a(n) ~ a(n-1) + a(n-2) + (Pi^2/(3*n) - 1)*a(n-3). For example, substituting the values a(47) = 124,754, a(48) = 147,273 and a(49) = 173,525 into the recurrence gives the approximation a(50) ~ 204,252.48... compared with the true value a(50) = 204,226. (End)

a(n) = Sum_{k=-inf..+inf} (-1)^k a(n-k(3k-1)/2) with a(0)=1 and a(negative)=0. The sum can be restricted to the (finite) range from k = (1-sqrt(1-24n))/6 to (1+sqrt(1-24n))/6), since all terms outside this range are zero. - Jos Koot, Jun 01 2016